If x, y satisfy x − y + 5 ≥ 0 x + y ≥ 0 x ≤ 3, then the minimum value of 3x + 4Y is () A. 52B. -10C. 0D. -3

If x, y satisfy x − y + 5 ≥ 0 x + y ≥ 0 x ≤ 3, then the minimum value of 3x + 4Y is () A. 52B. -10C. 0D. -3

By making the feasible region corresponding to the inequality system (as shown in the shadow) and moving the straight line, we can see that when the objective function z = 3x + 4Y passes through point a (3, - 3), the minimum value is taken, so the minimum value of 3x + 4Y is: 3 × 3 + 4 × (- 3) = - 3, so D is selected